integral from 0 to 2 of 2pix(6e^{-x^2})

Lösung

\int_{0}^{2} 2 π x (6 e^{- x^{2}}) d x

6 π (1 - \frac{1}{e ^{4}})

Dezimale

18.50431 \dots

Schritte zur Lösung

\int_{0}^{2} 2 π x \cdot 6 e^{- x^{2}} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 π 6 \cdot \int_{0}^{2} x e^{- x^{2}} d x

Vereinfache

= 12 π \cdot \int_{0}^{2} x e^{- x^{2}} d x

Wende U-Substitution an

= 12 π \cdot \int_{0}^{- 4} - \frac{e ^{u}}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 12 π (- \int_{- 4}^{0} - \frac{e ^{u}}{2} d u)

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 12 π (- (- \frac{1}{2} \cdot \int_{- 4}^{0} e^{u} d u))

Nutze das gemeinsame Integral :

\int e^{u} d u = e^{u}

= 12 π (- (- \frac{1}{2} [e^{u}]_{- 4}^{0}))

Vereinfache

12 π (- (- \frac{1}{2} [e^{u}]_{- 4}^{0})) : 6 π [e^{u}]_{- 4}^{0}

= 6 π [e^{u}]_{- 4}^{0}

Berechne die Grenzen:

1 - \frac{1}{e ^{4}}

= 6 π (1 - \frac{1}{e ^{4}})

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